Problem: Solve for $x$ and $y$ using elimination. ${3x-y = 10}$ ${4x+y = 32}$
Explanation: We can eliminate $y$ by adding the equations together when the $y$ coefficients have opposite signs. Add the equations together. Notice that the terms $-y$ and $y$ cancel out. $7x = 42$ $\dfrac{7x}{{7}} = \dfrac{42}{{7}}$ ${x = 6}$ Now that you know ${x = 6}$ , plug it back into $\thinspace {3x-y = 10}\thinspace$ to find $y$ ${3}{(6)}{ - y = 10}$ $18-y = 10$ $18{-18} - y = 10{-18}$ $-y = -8$ $\dfrac{-y}{{-1}} = \dfrac{-8}{{-1}}$ ${y = 8}$ You can also plug ${x = 6}$ into $\thinspace {4x+y = 32}\thinspace$ and get the same answer for $y$ : ${4}{(6)}{ + y = 32}$ ${y = 8}$